PY1001 Problem Set 10 – Solutions
(1) What is the minimum amount of energy in joules required to completely
melt 130 g of silver initially at 15.0◦ C?
The specific heat of silver is c = 236 J/kg K, the melting point of silver is 1235 K, and the
latent heat of fusion of silver is LF = 1.05 × 105 J/kg. So first, we must heat the silver to its
melting point (Q1 ), then melt it (Q2 ):
Q1 = mc∆T = (0.130 kg)(236 J/kg K)(1235 K − 288 K) = 2.90 × 104 J
Q2 = mL = (0.130 kg)(1.05 × 105 J/kg) = 1.36 × 104 J
The total heat required to heat and melt the silver is
Q = 2.90 × 104 J + 1.36 × 104 J = 4.26 × 104 J = 42.6 kJ
(2) An insulated thermos contains 130 cm3 of hot coffee at a temperature of
80.0◦ C. You put in 12.0 g of ice at its melting point to cool the coffee. By how
many degrees has your coffee cooled just after all the ice has melted? Treat the
coffee like water and assume that the thermos doesn’t let any heat in or out.
Equate the heat absorbed by the ice as it melts with the heat leaving the coffee as it cools:
mcof f ee cwater ∆T = mice LF,ice
∆T =
mice LF,ice
mcof f ee cwater
The specific heat of water is cwater = 4187 J/kg K, and the latent heat of fusion of ice is
LF,ice = 3.33 × 105 J/kg.
The density of water is ρ = 1 g/cm3 . Therefore, the mass of the coffee is mcof f ee = ρV =
(1 g/cm3 )(130 cm3 ) = 130 g = 0.130 kg.
∆T =
(0.012 kg)(3.33×105 J/kg)
(0.130 kg)(4187 J/kg K)
∆T = 7.34 K = 7.34◦ C −→ The coffee has cooled by 7.34◦ C.
TC
H
TH
L
W
(3) The slab has L = 25.0 cm and width W = 30.0 cm and is made of copper.
TH = 125◦ C, TC = 10.0◦ C, and the rate of heat conduction through the slab is
1660 J/s when a steady state (equilibrium) has been reached. (a) What is the
height of the slab? The same slab of copper is now at a temperature of T = 50 ◦ C
and is suspended in a oven that has a temperature of 200 ◦ C. What is the rate
at which the slab (b) emits and (c) absorbs thermal radiation if its emissivity is
= 0.85?
(a) The distance between the hot and cold ends of the slab is L = 0.25 m, and the area
through which the heat flows is A = W H = 0.30H m2 . The thermal conductivity of copper
is k = 400 W/m K.
C
C
Pcond = kA TH −T
= kW H TH −T
L
L
H=
Pcond L
kW (TH −TC )
H=
(1660 J/s)(0.250 m)
(400 W/m K)(0.30 m)(125−10)K
= 0.030 m = 3.0 cm
(b) The rate at which the slab emits radiation is
4
Prad = σATslab
where A is the total surface area of the slab and Tslab = 323 K. The total surface area of the
slab is
A = 2 × (W L) + 2 × LH + 2 × W H = 2(W L + LH + W H)
A = 2[(0.30 m)(0.25 m) + (0.25 m)(0.030 m) + (0.30 m)(0.030 m)] = 0.183 m 2
4
Prad = σATslab
= (5.67 × 10−8 W/m2 K4 )(0.85)(0.183 m2 )(323 K)4 = 96.0 W
(c) The rate at which the slab absorbs radiation is
4
Pabs = σAToven
where A is the total surface area of the slab and Toven = 473 K.
4
Pabs = σAToven
= (5.67 × 10−8 W/m2 K4 )(0.85)(0.183 m2 )(473 K)4 = 441 W
(4) The density of a gas is 1.24 × 10−2 kg/m3 at a temperature of 273 K and a
pressure of 1.00×10−2 atm. (a) Show that the molecular mass of the gas is given by
m = NρVA n , where V is the volume occupied by the gas and n is the number of moles
of the gas. (b) Use the result of (a) and the information given in the problem to
find the molecular mass of the gas and identify the gas. (c) What would be the
volume occupied by one mole of the gas at standard temperature and pressure?
(a) The molecular mass will be the total mass divided by the number of molecules present in
the gas:
m = number mass
= NρVA n
of molecules
(b) The ideal gas law says that
P V = nRT −→ Vn = RT
P
But the result of (a) shows that
V
A
A
= mN
−→ mN
= RT
n
ρ
ρ
P
Solving for m:
m=
ρRT
NA P
=
(1.24×10−2 kg/m3 )(8.31 J/mol K)(273 K)
(6.02×1023 )(1.01×103 Pa)
−26
m = 4.62 × 10
kg
The mass of a proton is mp = 1.67 × 10−27 kg. This means that there are
4.62×10−26 kg
1.67×10−27 kg
≈ 28
particles with the mass of the proton in each molecule. Since each nucleus will have approximately equal numbers of protons and neutrons, the gas could be made up of atoms with 14
protons in each nucleus – this would be silicon. But silicon is not a gas at such temperatures.
If there are two atoms per molecule, then the gas would be nitrogen, which has 7 protons and
7 neutrons in each nucleus. Thus, the gas is nitrogen.
(c) Use the ideal gas law:
P V = nRT −→ V = nRT
P
V =
(1 mol)(8.31 J/mol K)(273 K)
1.01×103 Pa
= 2.24 m3 .